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Finding higher order Derivatives

  1. Nov 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Let p be an arbitrary polynomial
    p(x) = anxn + an-1xn-1 + ... + a1x + a0, an cannot equal 0.
    a) Find (dn/dxn)[p(x)]
    b)What is (dk/dxk)[p(x)] for k>n


    2. Relevant equations



    3. The attempt at a solution
    Im actually not really sure what to do for this question.

    For the first part i tried to take the first three derivatives and got

    (d/dx)[p(x)] = a1
    (d2/dx2)[p(x)] = 2a2x + a1
    (d3/dx3)[p(x)] = 3a3x2 + 2a2x + a1

    (dn/dxn)[p(x)] = nanxn-1 + (n-1)a(n-1)xn-2 + ... + a1

    Thats my guess, im not sure of how to do it.
    As for part b:
    I dont even know how to approach this question, my only assumption is that its the same answer again with all the n's replaced with k's. I am certain this is wrong so please help me do this question both a and b. Thank you.
     
  2. jcsd
  3. Nov 1, 2011 #2

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    Welcome to PF, tmlrlz! :smile:

    What you have calculated is d/dx in each case.
    dn/dxn means that you take the derivative n times.
     
  4. Nov 1, 2011 #3
    I thought i did that for part a, did i do it wrong? And how would you go about doing part b, what is it asking and how is it different from part a? Thank you.
     
  5. Nov 1, 2011 #4

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    Let me give you an example:

    You wrote:
    (d2/dx2)[p(x)] = 2a2x + a1

    But what you actually did was:
    (d/dx)[p(x)] = 2a2x + a1

    It should be:
    (d2/dx2)[p(x)] = 2a2

    Can you do this for say n=3?



    If you understand what is intended for part a, I think you'll see what is intended for part b.
     
  6. Nov 1, 2011 #5
    The difference is that the n in the first part is the degree of the polynomial. The k in the second part is just any integer greater than the degree of the polynomial.
     
  7. Nov 1, 2011 #6

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    Yes, this is true.
    So I guess you answered the last part of your own question.
     
  8. Nov 1, 2011 #7

    oh i can't believe that i forgot to do the rest of the derivative!
    so for n = 3
    (d3/dx3)[p(x)] = 6a3

    And then you can find part a
    (dn/dxn)[p(x)] = n! * (an)

    However i am still confused as to what part b is requiring?
     
  9. Nov 1, 2011 #8

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    Right! :smile:

    Okay, let's pick a k>n.
    I'll pick k=n+1.

    What is (dn+1/dxn+1)[p(x)]?
     
  10. Nov 1, 2011 #9
    So then that means that when n = 1 we are taking the derivative of d2/dx2 right?

    I took the first 3 derivatives again
    when n =1
    d2/dx2[p(x)] = d/dx (a1) = 0
    when n =2
    d3/dx3[p(x)] = 0
    when n =3
    d4/dx4[p(x)] = 0

    so that means that
    (dk/dxk)[p(x)] = 0

    right? i hope!
     
    Last edited by a moderator: Nov 1, 2011
  11. Nov 1, 2011 #10

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    Yep! :wink:

    What do you think will happen if we pick k=n+2?
    Or k=n+3?
    Or any k>n?
     
  12. Nov 1, 2011 #11
    Then you will keep taking the derivative of zero which is zero over and over again till the end of the derivative instead of the previous case where you only got zero once, instead it will occur multiple times for that specific derivative
     
  13. Nov 1, 2011 #12

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    So? Any other questions?
     
  14. Nov 1, 2011 #13
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